How to derive the formula for $\sum_{n=0}^{\infty}\tan^{-1}(\frac{x^{2}}{n^{2}})$

Question

How to derive the formula $$\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{x^{2}}{n^{2}}\right)=\tan^{-1}\left[\frac{\tan\left(\frac{x\pi}{\sqrt{2}}\right)-\tanh\left(\frac{x\pi}{\sqrt{2}}\right)}{\tan\left(\frac{x\pi}{\sqrt{2}}\right)+\tanh\left(\frac{x\pi}{\sqrt{2}}\right)}\right]$$

Is the formula correct? How do I derive the above formula? I tried to apply the formula integral as the limit of sum formula.

Answer 1

I'd start that sum at $n=1$ if I were you.

Now, up to multiples of $\pi$, $$\sum_{n=1}^\infty\tan^{-1}\frac{x^2}{n^2} $$ is the argument of the complex number $$\prod_{n=1}^\infty\left(1+\frac{ix^2}{n^2}\right).$$

But $$\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)=\frac{\sin\pi z}{\pi z}$$ so how about choosing $z$ to make $-z^2=ix^2$?